Calculus Question

Question

If f(x) = sinx, evaluate f(2+h)-f(2)/h to two decimal places as h approaches 0. Help?!

Answer 1

Hi John,

 

In order to do this problem using the limit of a difference quotient you need to be aware of two theorems on limits involving sin and cos. 

 

1. lim_t->0 (sin t)/t  = 1

 

2. lim_t->0 (1 - cos t)/t = 0

 

Assuming you have those theorems available for your use:

 

 lim_h->0(sin(2+h)-sin(2))/h 

 

At this point use the trig identity for the sum of two angles to expand the sin(2+h) = sin2cosh + cos2sinh.

 

lim_h->0 (sin2cosh + cos2sinh - sin2)/h: rearrange terms in the numerator and separate into separate fractions (with h in the denominator)

 

lim_h->0 (sin2cosh-sin2)/h  + lim_h->0(cos2sinh)/h

 

At this point we can do something like this:

 

lim_h->0 sin2 (cosh-1)/h  + lim_h->0 cos2 (sinh)/h

 

The last two factors on either side of the + sign are where the two limit theorems are applied. 

 

Without being overly precise we get something like:

 

sin2 (0) + cos2 (1) = cos 2.  At that point evaluate the cos2 using a caculator where your calculator is in radian mode. 

 

Eventually you will learn that the limit of a difference quotient is the derivative of the given function. The above specific problem is that pattern that is used to prove that the derivative of sin x = cos x.

Answer 2

It is easier to realize that

 

lim       f(x + h) - f(x)

h->0   ____________   =  f'(x)

                    h

 

lim       f(2 + h) - f(2)  f'(2)   

h->0   _________________  = f'(2) = cos(2)

                       h